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Ta có
arctanxx2=−xx2+1+1x−xx2+1−arctanxx2
arctanxx2=(−12ln(x2+1))′+(lnx)′−(arctanxx)′
arctanxx2=(lnx√x2+1)′−(arctanxx)′
Suy ra
+∞∫1arctanxx2dx=[lnx√x2+1−arctanxx]+∞1=π+ln44
Chú ý rằng
lim
\lim_{x \to +\infty}\dfrac{\arctan x}{x} \underbrace{=}_{L'Hospital}\lim_{x \to +\infty}\dfrac{(\arctan x)'}{(x)'}=\lim_{x \to +\infty}\dfrac{\dfrac{1}{x^2+1}}{1} =0
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